Question

If the temp. of 63.4 grams of water changes from 22.3C to 28.7C, how many joules of heat are involved? Show work

Answer 1

Q = 1698.51 J

Step-by-step explanation:

Given that,

Mass of water, m = 63.4 grams

It changes temperature from 22.3C to 28.7C.

We need to find the heat involved. The heat involved due to the change in temperature is given by :

`Q=mc\Delta T`

Where

c is the specific heat of water, c = 4.186 joule/gram °C

So,

`Q=63.4 * 4.186 * (28.7- 22.3)\\\\Q=1698.51\ J`

So, 1698.51 J of heat is involved.